1. Irreducible Scheme A topological space X is irreducible if X is nonempty and can not be a union of two proper closed subsets. A nonempty closed subset Z of X is irreducible if Z equipped with the induced topology from X is an irreducible topological space. A point ξ of an irreducible closed subset Z of X is called a generic point if Z = {ξ}. Proposition 1.1. A topological space X is irreducible if and only if the intersection of any two nonempty open subsets of X is nonempty. Proof. Suppose that X is irreducible. Assume that there exist two nonempty open subsets U1,U2 so that U1 ∩ U2 = ∅. Denote Zi = X \ Ui for i = 1, 2. Then Z1 and Z2 are proper closed subsets of X. Since U1 ∩ U2 = ∅,Z1 ∪ Z2 = X. This leads to a contradiction that X is irreducible. Hence any two nonempty open subsets of X is nonempty. Conversely, suppose that the intersection of any two nonempty open subsets of X is nonempty. Assume that X = Z1 ∪ Z2 is a union of closed subsets of X. Denote Ui = X \ Zi for i = 1, 2. Then U1 ∩ U2 = ∅. This shows that either U1 = ∅ or U2 = ∅. This is equivalent to say that Z1 = X or Z2 = X. Hence X is irreducible.  Definition 1.1. A scheme is irreducible if its underlying topological space is irreducible. Lemma 1.1. Let X = Spec A be the spectrum of a ring A and D(f) be the distinguished open set associated with f ∈ A. Then D(f) = ∅ if and only if f is nilpotent. Proof. Let f be nilpotent. Then f n = 0 for some n > 0. In other words, f n ∈ p for all p ∈ X. Since p is prime, f ∈ p. We see that f ∈ p for all p ∈ X. Hence p ∈ V (f) for all p ∈ X. We obtain X = V (f). Hence D(f) = ∅. Conversely, suppose D(f) = ∅. Then X = V (f). This shows that f ∈ p for all p. Hence T f ∈ p∈X p = Nil(A). (The nilradical of A.) Hence f is nilpotent.  Proposition 1.2. Let X = Spec A be the spectrum of a ring A. Then X is irreducible if and only if its nil radical Nil(A) is a prime ideal. Proof. Let us assume that ξ = Nil(A) is a prime ideal. Then the Nil(A) is the intersection of all prime ideals of A. Hence p ∈ V (ξ) for all p ∈ X. Hence X = V (ξ). Notice that V (ξ) is a closed subset of X containing ξ. Then V (ξ) contains the closure of ξ. Moreover, if V (I) is a closed subset of X containing ξ, then ξ ⊃ I. Thus V (ξ) ⊂ V (I). This implies that V (ξ) is in fact the closure of ξ. In other words, ξ is the generic point. Hence all nonempty open subsets of X contain ξ. Hence the intersection of any two nonempty open subsets of X contains ξ. Suppose that X is irreducible. Let fg ∈ Nil(A). To show that Nil(A) is a prime, we need to show that either f ∈ Nil(A) or g ∈ Nil(A). Then D(fg) = D(f)∩D(g). If both D(f) and D(g) are nonempty, then f and g are not nilpotent. Since X is irreducible and D(f) and D(g) are nonempty open subsets of X,D(f) ∩ D(g) = D(fg) is nonempty. This implies that fg is not nilpotent, i.e. fg 6∈ Nil(A). This leads to the contradiction to the assumption that fg ∈ Nil(A). Thus either D(f) = ∅ or D(g) = ∅. Hence either f or g is nilpotent, i.e. f ∈ Nil(A) or g ∈ Nil(A).  Proposition 1.3. Let X = Spec A. For p ∈ Spec A, V (p) is an irreducible closed subset. Conversely, every irreducible closed subset of X is of the form V (p) for some p ∈ X. 1 2

Proof. Let Z = V (I) be a nonempty closed subset of X where I is a radical ideal of X. Suppose I is not prime. Then we can choose f, g ∈ A so that fg ∈ I but f, g 6∈ I. In this case, we consider (f) + I and (g) + I. Then V ((f) + I) and V ((g) + I) are proper closed subsets of V (I). Moreover, V ((f) + I) ∪ V ((g) + I) = V ((fg) + I) = V (I) = Z. This implies that Z is not irreducible. If Z is not irreducible, Z is a union of two proper closed subsets V (I1) and V (I2). Assume that I1 and I2 are radical ideals of A. Then I1 ⊃ I and I2 ⊃ I. Choose fi ∈ Ii \ I. Then (fi) + I ⊃ I for i = 1, 2. Then V ((fi) + I) ⊃ V (Ii). We see that

V ((f1f2) + I) = V ((f1) + I) ∪ V ((f2) + I) = V (I). p p Then (f1f2) + I = I. Since (f1f2) ⊂ (f1f2) + I ⊂ (f1f2) + I, we see that f1f2 ∈ I. This shows that I is not a prime.  If p is a prime ideal of a ring A, then V (p) is the closure of p in Spec A. Proposition 1.4. Let X be a scheme. Suppose X is irreducible. Then X has a unique generic point. Proof. Suppose X = Spec A. We have seen that ξ = Nil(A) is the unique generic point. Since X is irreducible, any affine open subset U of X is also irreducible. Suppose X (not necessarily affine) has two generic points ξ and η. Then ξ and η are also generic points of U of X. Since U is affine, ξ = η by the uniqueness of the generic point of an affine scheme. Hence we proved the uniqueness of generic point of a scheme. Now let us prove the existence. The affine open subsets of X forms a basis for the Zariski topology of X. Since X is irreducible, the intersection of any two nonempty affine open subsets of X are nonempty. Hence there exists a nonempty affine open set contained in the intersection of any two nonempty affine open subsets of X. Let U, V be nonempty affine open sets. Choose ξ and η so that ξ and η are generic point of U and V respectively. Choose W nonempty affine open so that W ⊂ U ∩ V. Then we know both ξ and η are generic point of W. By uniqueness of the generic point of W, ξ = η. This proves that all the affine open subsets of X share the same generic point ξ. Now let us prove that ξ is the generic point of X. Let x be any point of X and U be any open neighborhood of x. Since X is a scheme, we can choose an affine open neighborhood V of x contained in U. Since ξ is a generic point of V,U contains ξ. Hence x lies in the closure of {ξ}. Thus X = {ξ}.  Proposition 1.5. Let X be a scheme. Any irreducible closed subset of X has a unique generic point. Proof. Let Z be an irreducible closed subset of X. For any affine open subset U of X, we consider the intersection U ∩ Z. Either U ∩ Z is empty or U ∩ Z is nonempty. Suppose that U ∩ Z is nonempty. Then U ∩ Z is an irreducible closed subset of U. Write U = Spec A. Then U ∩ Z = V (I) for some ideal I. Since U ∩ Z is irreducible, I = p is a prime ideal. Then V (p) is the closure of p in U ∩ Z. Hence p is the generic point of U ∩ Z. Let ξ be the generic point of U ∩ Z. Similar we can show that U ∩ Z share the same generic point ξ for all affine open subset U of X. Using the fact that affine open subsets forms a base for the Zariski topology of X, ξ is the generic point of Z. (The detailed proof of these statements is the same as that of the above proposition.)  3

Proposition 1.6. Let X be a scheme. T.F.A.E. (1) X is irreducible. (2) There exists an affine open covering {Ui} so that Ui is irreducible and Ui ∩ Uj 6= ∅ for all i, j. (3) X is nonempty and every nonempty affine open U ⊂ X is irreducible. Proof. (1) implies (2). Since X is irreducible, it has a unique generic point ξ. Since X is a scheme, we can choose an affine open covering {Ui} for X so that Ui are all nonempty. Since Ui is nonempty, Ui contains the generic point ξ for all i. This implies that Ui ∩ Uj contains ξ and hence is nonempty. Moreover any subspace of irreducible topological space is also irreducible. Hence Ui is irreducible. Similarly, (1) implies (3). (2) implies (1). Let X = Z1 ∪ Z2 where Z1 and Z2 are closed subsets of X. For any Ui, either Ui ∩ Z1 = ∅ or Ui ∩ Z2 = ∅. Hence either Ui ⊂ Z1 or Ui ⊂ Z2. We may assume that there exists Ui so that Ui ⊂ Z1. Since Ui is irreducible, Ui ∩ Z1 = Ui is an irreducible closed subset of Ui. Then Ui ∩ Z1 has a unique generic point ξ. For any Uj, we know that Ui ∩ Uj is nonempty by assumption and hence contains ξ. Then Ui ∩ Uj is densed in Uj, and contained in Z1 ∩ Uj. Notice that Z1 ∩ Uj is a closed set containing ξ, then Z1 ∩ Uj contains the closure of ξ in Uj. Since Ui ∩ Uj is dense in Uj, the closure of ξ in Uj is Uj. S Hence Uj ⊂ Z1 ∩ Uj ⊂ Uj. We conclude that Uj ⊂ Z1. Hence X = i Ui ⊂ Z1 which implies that X = Z1. Thus X is irreducible. (3) implies (2). Since X is nonempty, we can choose an affine open covering {Ui} of X so that Ui are all nonempty. By assumption, Ui are all irreducible. Now we want to show that ` Ui ∩Uj 6= ∅. Suppose Ui ∩Uj is an empty set. Since Ui and Uj are affine, so is Ui Uj. Since ` both Ui and Uj are nonempty, Ui Uj is also nonempty. Since every affine open subset of ` X is irreducible, Ui Uj is irreducible but this is impossible. Hence Ui ∩ Uj is nonempty for any i, j.